POJ-3140 Contestants Division （树）

题目大意：一棵树，带点权。将这棵树分成两部分，找出使得两部分的点权和的差最小。

题目分析：直接dfs即可。找出每棵子树u的点权和size(u)，如果以u和它的父节点之间的边为界，那么两边的点权和分别为sum-size(u)和size(u)。

代码如下：

# include<iostream>
# include<cstdio>
# include<cstring>
# include<vector>
# include<queue>
# include<list>
# include<set>
# include<map>
# include<string>
# include<cmath>
# include<cstdlib>
# include<algorithm>
using namespace std;
# define LL long long

const int N=1005;
const int INF=1000000000;
const LL oo=0x7fffffffffffffff;

struct Edge
{
	int to,nxt;
};
Edge e[N*200];
int n,m,cnt;
int head[N*100];
int w[N*100];
LL sum;
LL size[N*100];
int vis[N*100];

void add(int u,int v)
{
	e[cnt].to=v;
	e[cnt].nxt=head[u];
	head[u]=cnt++;
}

void init()
{
	cnt=0;
	memset(vis,0,sizeof(vis));
	memset(head,-1,sizeof(head));
	sum=0;
	for(int i=1;i<=n;++i){
		scanf("%d",w+i);
		sum+=(LL)w[i];
	}
	int a,b;
	for(int i=0;i<m;++i){
		scanf("%d%d",&a,&b);
		add(a,b);
		add(b,a);
	}
}

void dfs(int u,int fa)
{
	size[u]=w[u];
	vis[u]=1;
	for(int i=head[u];i!=-1;i=e[i].nxt){
		int v=e[i].to;
		if(v==fa||vis[v]) continue;
		dfs(v,u);
		size[u]+=size[v];
	}
}

LL solve()
{
	dfs(1,-1);
	LL ans=oo;
	for(int i=1;i<=n;++i){
		LL t=max(sum-size[i],size[i])-min(sum-size[i],size[i]);
		if(t<ans) ans=t;
	}
	return ans;
}

int main()
{
	int cas=0;
	while(scanf("%d%d",&n,&m)&&(n+m))
	{
		init();
		printf("Case %d: %lld
",++cas,solve());
	}
	return 0;
}